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\oint _{C}(\sin zdx-\cos xdy+\sin ydz)\,

=\oint _{C}(\sin i-\cos xj+\sin yk)\cdot (dxi+dyj+dzk)=\oint _{C}F\cdot dr\, --(1)

where F=(\sin zi-\cos xj+\sin yk)\, --(2)

By Stokes'theorm,\oint _{C}F\cdot dr=\iint _{S}{\mathrm  {curl}}F\cdot ndS\, --(3)

{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\\sin z&-\cos x&\sin y\end{vmatrix}}=\cos yi+\cos zj+\sin xk\,

Here the rectangle S given by 0\leq x\leq \pi ,0\leq y\leq 1\, lies in the plane z=3 and so the unit vector normal to S is k. Thus,here n=k.

Therefore,{\mathrm  {curl}}F\cdot n=(\cos yi+\cos zj+\sin xk)\cdot k=\sin x\, --(4)

So,from (1),(3) and (4),we have \oint _{C}(\sin zdx-\cos xdy+\sin ydz)\,

=\int _{{y=0}}^{{1}}\int _{{x=0}}^{{\pi }}\sin xdxdy=\int _{0}^{1}[-\cos x]_{{0}}^{{\pi }}dy\,

=2\int _{0}^{1}dy=2\,

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